这是一份nottingham诺丁汉大学MATH1007作业代写的成功案例
$$
\hat{L} \hat{U}=\hat{P}(A+\delta A) \quad \text { with }|\delta A| \leq \frac{2 n \epsilon}{1-n \epsilon}|\hat{L}||\hat{U}|
$$
and for the particular case that $m=n$ and $A$ is nonsingular, if an approximate solution, $\hat{\mathbf{x}}$, to $A \mathbf{x}=\mathbf{b}$ is computed by solving the two triangular linear systems, $\hat{L} \mathbf{y}=\hat{P} \mathbf{b}$ and $\hat{U} \hat{\mathbf{x}}=\mathbf{y}$, then $\hat{\mathbf{x}}$ is the exact solution to a perturbed linear system:
$$
(A+\delta A) \hat{\mathbf{x}}=\mathbf{b} \quad \text { with } \quad|\delta A| \leq \frac{2 n \epsilon}{1-n \epsilon} \hat{P}^{T}|\mathcal{L}||\hat{U}|
$$
Furthermore, $\left|L_{i, j}\right| \leq 1$ and $\left|U_{i, j}\right| \leq 2^{i-1} \max {k \leq i}\left|A{k, j}\right|$, so
$$
|\delta A|_{\infty} \leq \frac{2^{n} n^{2} \epsilon}{1-n \epsilon}|A|_{\infty}
$$
MATH1007 COURSE NOTES :
$$
\hat{G} \hat{G}^{T}=A+\delta A \quad \text { with } \quad|\delta A| \leq \frac{(n+1) \epsilon}{1-(n+1) \epsilon}|\hat{G}|\left|\hat{G}^{T}\right| .
$$
Furthermore, if an approximate solution, $\hat{\mathbf{x}}$, to $A \mathbf{x}=\mathbf{b}$ is computed by solving the two triangular linear systems $\hat{G} \mathbf{y}=\mathbf{b}$ and $\hat{G}^{T} \hat{\mathbf{x}}=\mathbf{y}$, and a scaling matrix is defined as $\Delta=\operatorname{diag}\left(\sqrt{a_{i i}}\right)$, then the scaled error $\Delta(\mathbf{x}-\hat{\mathbf{x}})$ satisfies
$$
\frac{|\Delta(\mathbf{x}-\hat{\mathbf{x}})|_{2}}{|\Delta \mathbf{x}|_{2}} \leq \frac{\kappa_{2}(H) \epsilon}{1-\kappa_{2}(H) \epsilon}
$$